The sum $$\sum_{r=0}^n(n-r)^2r^2(n-2r)$$ is either equal to $$\sum_{k=0}^{n-1}e^{\frac{-2k\pi i}{n}} or \sum_{k=0}^{n}e^{\frac{2k\pi i}{n}}$$Which one is it?
Summation in euler's terms
1
$\begingroup$
combinatorics
complex-numbers
2 Answers
2
Hint:
Substituting $r\mapsto n-r$ gives $$ \begin{align} \sum_{r=0}^n(n-r)^2r^2(n-2r) &=\sum_{r=0}^n(n-r)^2r^2(2r-n)\\ &=-\sum_{r=0}^n(n-r)^2r^2(n-2r) \end{align} $$ Furthermore, $$ \sum_{k=0}^{n-1}e^{\frac{-2k\pi i}{n}}=\frac{e^{\frac{-2n\pi i}n}-1}{e^{\frac{-2\pi i}n}-1}\\ $$
0
HINT: $$(n-r)^2r^2(n-2r)=(n^2-nr)^2(n-2r)=n^5-4n^4r+5n^3r^2-2n^2r^3$$
$$\implies\sum_{r=0}^n(n-r)^2r^2(n-2r)=n^5\sum_{r=0}^n1-4n^4\sum_{r=0}^nr+5n^3\sum_{r=0}^nr^2-2n^2\sum_{r=0}^nr^3$$